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Yes. Take $G=F_2$. The free product satisfiesA connected three-sheeted cover of the two-petal rose has fundamental group of rank $1+3(2-1)=4$ by the Nielsen–Schreier formula. Thus $F_4$ is isomorphic to an index-three subgroup of $F_2$. A finite-index subgroup quasi-isometry then gives

Cover the attached disc by a slightly enlarged interior and a collar of its boundary, and apply the Seifert-van Kampen theorem. The collar intersection has fundamental group generated by the boundary loop, whose image in $X$ is $[\alpha ]$, while that loop becomes trivial in the disc. Van Kampen therefore quotients by the smallest normal subgroup containing $[\alpha ]$:This is the general fundamental group after attaching a 2-cell.

Start with a wedge of two circles, whose fundamental group is the free groupAttach a disc along the loop represented by $a^3{b}^{-7}$. The displayed quotient gives a connected two-dimensional CW complex $X$ with

Write the closed oriented surface of genus $g$ as ${\Sigma }_g=A\cup B$, where $B$ is a closed disk, $A$ is a slightly enlarged once-punctured surface, and $A\cap B$ is an annulus. Thus $B$ is contractible, $A\cap B$ is homotopy equivalent to the circle, and $A$ is homotopy equivalent to the wedge sum of $2g$ circles. The boundary circle represents the product of the $g$ commutators in the fundamental group of $A$, so its image in the first homology group, which is the abelianization of that group, is zero.

The relevant part of the Mayer–Vietoris sequence is consequentlyThe degree-zero part says that the connected space ${\Sigma }_g$ has $H_0({\Sigma }_g)\cong \mathbb{Z}$, while the groups above degree two vanish. Hence

Send each orientation-preserving isometry to its isotopy class. This is a group homomorphismFor $g\ge 2$, a hyperbolic isometry isotopic to the identity is the identity. One proof lifts it to the hyperbolic plane: after composing with a deck transformation, its lift commutes with the surface fundamental group. It consequently fixes the endpoints at infinity of every hyperbolic deck transformation. Those endpoints are dense in the circle at infinity, so the lift fixes that circle pointwise and is the identity. The homomorphism is therefore injective, proving the injection of a finite hyperbolic isometry group into a mapping class group.

Both low-genus analogues fail. Every orientation-preserving homeomorphism of the unit sphere is isotopic to the identity, but a round sphere has nontrivial finite rotation groups. On a flat torus, translation by a nonzero torsion point is a finite-order orientation-preserving isometry isotopic to the identity.

No. A conformal bijection is a homeomorphism and therefore induces an isomorphism of fundamental groups. The simply connected domain $D$ has trivial fundamental group, whereasThis contradiction rules out such a map.

For general odd $m_{+}$ and $m_{-}$, the same rectangular circuit givesEvery elementary two-dimensional nematic defect has $|q|=1/2$, so the minimum number is

In three dimensions the director takes values in the real projective plane, whose fundamental group is $\mathbb{Z}/2\mathbb{Z}$. The signs of half-charge line defects are no longer distinct topological classes, and two such lines can annihilate by escape into the third dimension. Hence only the parity of $N_{2D}$ remains:When one is required, it is a disclination line extending through the unbounded $z$ direction.

Suppose first that $F:D^n\to X$ extends $f$. Regarding $S^{n-1}$ as the boundary of the unit disc,is a homotopy from $f$ to the constant map with value $F(0)$. Thus $f$ is null-homotopic. Conversely, if $H:S^{n-1}\times [0,1]\to X$ is a homotopy from $f$ to a constant, then $H$ is constant on $S^{n-1}\times \{1\}$ and hence descends to the quotientwhich is the cone on $S^{n-1}$ and is homeomorphic to $D^n$. The descended map extends $f$. This proves the extension-null-homotopy criterion for a sphere.

A universal cover of $X$ is a covering map $p:\tilde{X}\to X$ whose total space is path-connected and simply connected. Let $p_1:({\tilde{X}}_1,{\tilde{x}}_1)\to (X,x)$ and $p_2:({\tilde{X}}_2,{\tilde{x}}_2)\to (X,x)$ be two universal covers. The lifting criterion for a covering space applies becauseso $p_1$ has a based lift $h:{\tilde{X}}_1\to {\tilde{X}}_2$ satisfying $p_{2}h=p_1$. Similarly there is a based lift $k:{\tilde{X}}_2\to {\tilde{X}}_1$ of $p_2$. Both $kh$ and the identity are lifts of $p_1$ that agree at the base point, so uniqueness of lifts gives $kh=1$. Likewise $hk=1$. Hence $h$ is a homeomorphism. This is the uniqueness of a universal covering space.

Now let $p:\tilde{X}\to X$ be universal with $\tilde{X}$ contractible. If an extension $F:|K|\to X$ exists, functoriality of the fundamental group givesThus the required factorization holds with $\Phi =F_{\ast }$.

Conversely, suppose $f_{\ast }=\Phi i_{\ast }$. We extend over the simplices of $K$ by dimension. The boundary of every 2-simplex is a loop in $|K^1|$ that becomes null-homotopic in $|K|$, so its class lies in $\ker i_{\ast }$. The factorization givesHence $f{|}_{\partial \sigma }$ is null-homotopic in $X$ and extends over the disc $|\sigma |$ by the first part. Since distinct 2-simplices meet along the already fixed 1-skeleton, these extensions combine to a map on $|K^2|$.

Inductively suppose the map is defined on $|K^{m-1}|$ for $m\ge 3$. For an $m$-simplex $\sigma$, its boundary is $S^{m-1}$, which is simply connected. The boundary map therefore lifts through $p$ by the covering-space lifting criterion. Its lift into the contractible space $\tilde{X}$ is null-homotopic, so the boundary map itself is null-homotopic and extends over $|\sigma |$. Extending over every $m$-simplex and then every skeleton constructs $F:|K|\to X$. The weak topology on a simplicial complex makes the cellwise map continuous. This proves the extension criterion into a space with contractible universal cover.

A based loop is a continuous map $\alpha :[0,1]\to X$ with $\alpha (0)=\alpha (1)=x_0$. Two loops are equivalent when they are joined by a homotopy that keeps both endpoints at $x_0$. The fundamental group ${\pi }_1(X,x_0)$ is the set of these based-homotopy classes.

For loops $\alpha$ and $\beta$, define concatenation byIf $H$ and $K$ are based homotopies from $\alpha$ to ${\alpha }^{\prime }$ and from $\beta$ to ${\beta }^{\prime }$, concatenating $H(s,-)$ and $K(s,-)$ gives a based homotopy from $\alpha \ast \beta$ to ${\alpha }^{\prime }\ast {\beta }^{\prime }$. Thus multiplication $[\alpha ][\beta ]=[\alpha \ast \beta ]$ is well-defined.

Concatenating three paths with different breakpoints changes only the speed of traversal. Linear interpolation between the two increasing piecewise-linear parametrizations gives a based homotopy, proving associativity on classes. The constant loop $c(t)=x_0$ is an identity, since deleting its stationary half is another endpoint-fixing reparametrization. The inverse of $[\alpha ]$ is represented by $\bar{\alpha }(t)=\alpha (1-t)$. Indeed, $\alpha \ast \bar{\alpha }$ contracts throughand the analogous contraction handles $\bar{\alpha }\ast \alpha$. Hence the operation satisfies all group axioms.

The Bonnet-Myers theorem states that if a complete connected $n$-dimensional Riemannian manifold satisfiesfor some $k>0$, thenIn particular, $M$ is compact and has finite fundamental group.

By the Hopf-Rinow theorem, points $p,q\in M$ are joined by a unit-speed length-minimizing geodesic $\gamma :[0,\ell ]\to M$. Choose a parallel orthonormal frame $E_1,\dots ,E_{n-1}$ normal to $T=\dot{\gamma }$ and setThe endpoint-vanishing fields $V_i$ arise from fixed-endpoint variations. Since $\gamma$ minimizes length, its Riemannian index form is nonnegative on each $V_i$. Summing the second variation of Riemannian arc length givesUsing the Ricci curvature bound and integrating ${\sin }^2$ and ${\cos }^2$ yieldsso $\ell \le \pi /\sqrt{k}$. Taking the supremum over $p,q$ proves the diameter bound. Hopf-Rinow now makes the closed bounded space $M$ compact. Finally, the same bound applies to the complete universal cover; a compact universal cover has finite fibres over $M$, so ${\pi }_1(M)$ is finite.

A simply connected space is path connected and has trivial fundamental group: every closed loop contracts to a point. The three-sphere is simply connected, whereas Real projective space ${\mathbb{RP}}^3$ has fundamental group ${\mathbb{Z}}_2$.

For positive $p,q$, quotient $S^p\times S^q$ by $(x,y)\mapsto (-x,-y)$. This is a free double covering space action. Its closed manifold quotient is orientable exactly when $p+q$ is even, because the product mapping degree is $(-1{)}^{p+q+2}$. If $p,q\ge 2$, the product is simply connected and the quotient has fundamental group $\mathbb{Z}/2$. For $p=q=2$, the quotient has integral cohomology groups $\mathbb{Z}$ in degrees $0,4$, $\mathbb{Z}/2$ in degrees $2,3$, and zero elsewhere. Every positive-degree cup product is zero. These conclusions follow from Poincare duality, the Euler characteristic under a finite covering, and the universal coefficient theorem for cohomology.

The Adjoint double cover from SU(2) to SO(3) has kernel $\{I,-I\}$. Under SU(2) as the three-sphere, multiplication by $-I$ is the antipodal map, so the quotient is Real projective space ${\mathbb{RP}}^3$. Equivalently, an axis-angle ball of radius $\pi$ has opposite boundary points identified. The fundamental group is ${\mathbb{Z}}_2$.

For a connected space, this is the connected covering corresponding to the kernel of the homomorphism from its fundamental group to its abelianization. Its deck transformation group is $H_1(Y;\mathbb{Z})$. Lifted cells make its cellular chain complex a complex over the group ring $\mathbb{Z}[H_1(Y)]$.

The wedge of two circles is a graph with one vertex and two loop edges. Its fundamental group is the free group on the two oriented loops.